0 ___PreliminaryThis K. Pearson´s test (Goodness of Fit) is pointed as historically the first one to achieve the goal H0: data X and F(x) are not incompatible?The algorithm starts by forecasting, from the Distribution F(x), Model j (1>=j<=k), the sample items content, for each one of the k same amplitude intervals (bins) data is divided. In spite k be in principle arbitrary, statisticians had noted that the bins contents, Observed j, should not be less than 5 or 6 in order that the sample statistics:Q = Sum (Observed j - Model j)^2/Model j,do follow reasonably well the theoretical Chi - Squared Distribution when H0 is not inraisonable odd.

1 ___New approachLet be n the sample under analysis size. And suppose that we set each interval the same content n/k, which contrary to the classic way is not conditioned to a minimum: it could be any value. By other words: the interval amplitudes are adjusted such that is constant, k/n, their probability to contain the items. An even more important feature of the proposed algorithm is that we do not depend on any Distribution to reach the decision: the p-value follows directly from Monte Carlo simulations. Therefore we define the Ij= (aj, bj) intervals such that we have F(bj) - F(aj) = k/n, possibily a1= -infinity, bk= +infinity.

2___Application exampleTwo-hundred years ago, from the XVIII end to the beggining XIX siecle, Henry Cavendish (1731-1810) reported 23 measurements of the mean density of Earth, 5.488 g/cm3; today´s value is 5.513/gcm3 which differs from 0.45 % only.We will search, by the Chi squared test, if the former data could follow a Normal Distribution. Following the usual method we estimate the mean m= 5.488 and standard deviation s, then reduction data by Z(i)=(X(i)-m)/s. Based on k=5 the intervals, 0.2 probability each, are:_____(-infinity, -0.842)_____4 values_____bin 1_____(-0.842, -0.253)_______ 7_____________ 2_____(-0.253, 0.253)_______ 3_____________ 3_____( 0.253. 0.842 )_______ 4_____________ 4_____( 0.842, +infinity)_____5_____________ 5Q0= 2.000 and prob(Q>Q0) = 0.632= p-value with 5-2-1=2 Chi-square 2 degrees of freedom from NET software.A more correct p-value can be obtained by Monte Carlo simulations as follows (see Program CAV-DISH). From the statement c=int(RND*23/5)+1 we obtain with the same probability one of the 5 values c= 1, 2, 3, 4, 5 indicating the bin the value will fall. Performing 23 times we got a *vector* n1 to n5 which provide the simulated Q relative to the sample. This is the current sample which is repeated M times: the fraction of Q´s larger than the source sample Q0=2.000 do provide the exact p-value. From 4 repetitions (M=1 million each) we find 0.644, 0.644, 0.643, 0.644.Conclusion: Both methods leads to not disapprove that density data is Normal, at the usual alpha = 0.05. The main feature attached to the Monte Carlo one here presented is that, surely, Chi-Square Distribution approximate methodology can be easily avoided which is particularly useful when alpha and p-value do tend to be equal.